The last digit of `3^(40)` is

Options

A. 1
B. 3
C. 7
D. 9

Answer

To find the last digit of \(3^{40}\), we can use the concept of cyclicity of the last digits of powers of 3. Here’s a step-by-step solution:

### Step 1: Identify the Pattern of Last Digits
First, let's calculate the last digits of the first few powers of 3:
- \(3^1 = 3\) (last digit is 3)
- \(3^2 = 9\) (last digit is 9)
- \(3^3 = 27\) (last digit is 7)
- \(3^4 = 81\) (last digit is 1)
- \(3^5 = 243\) (last digit is 3)

From this, we can see that the last digits repeat every 4 powers:
- \(3, 9, 7, 1\)

### Step 2: Determine the Cyclicity
The cycle of last digits is:
- \(3^1 \rightarrow 3\)
- \(3^2 \rightarrow 9\)
- \(3^3 \rightarrow 7\)
- \(3^4 \rightarrow 1\)

This cycle repeats every 4 terms.

### Step 3: Find the Position in the Cycle
To find the last digit of \(3^{40}\), we need to determine which position \(40\) corresponds to in the cycle. We do this by calculating \(40 \mod 4\):
\[ 40 \div 4 = 10 \quad \text{(remainder 0)} \]
This means \(40\) is a multiple of \(4\), which corresponds to the last digit of \(3^4\).

### Step 4: Identify the Last Digit
From our cycle, we know that:
- The last digit of \(3^4\) is \(1\).

Thus, the last digit of \(3^{40}\) is also \(1\).

### Final Answer
The last digit of \(3^{40}\) is \(1\).

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